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Ilya [14]
3 years ago
7

Please help me with this physics prooblem

Physics
1 answer:
zaharov [31]3 years ago
5 0

Take the missile's starting position to be the origin. Assuming the angles given are taken to be counterclockwise from the positive horizontal axis, the missile has position vector with components

x=v_0\cos20.0^\circ t+\dfrac12a_xt^2

y=v_0\sin20.0^\circ t+\dfrac12a_yt^2

The missile's final position after 9.20 s has to be a vector whose distance from the origin is 19,500 m and situated 32.0 deg relative the positive horizontal axis. This means the final position should have components

x_{9.20\,\mathrm s}=(19,500\,\mathrm m)\cos32.0^\circ

y_{9.20\,\mathrm s}=(19,500\,\mathrm m)\sin32.0^\circ

So we have enough information to solve for the components of the acceleration vector, a_x and a_y:

x_{9.20\,\mathrm s}=\left(1810\,\dfrac{\mathrm m}{\mathrm s}\right)\cos20.0^\circ(9.20\,\mathrm s)+\dfrac12a_x(9.20\,\mathrm s)^2\implies a_x=21.0\,\dfrac{\mathrm m}{\mathrm s^2}

y_{9.20\,\mathrm s}=\left(1810\,\dfrac{\mathrm m}{\mathrm s}\right)\sin20.0^\circ(9.20\,\mathrm s)+\dfrac12a_y(9.20\,\mathrm s)^2\implies a_y=110\,\dfrac{\mathrm m}{\mathrm s^2}

The acceleration vector then has direction \theta where

\tan\theta=\dfrac{a_y}{a_x}\implies\theta=79.2^\circ

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Answer:

Explanation:

AB = 110 miles

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Answer:

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Explanation:

Using the formula for an electric field;

E = kQ/r²

k = 1/(4πε₀) = 8.99 × 10⁹ Nm²/C²

E = 100N/C

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Q = Er²/k = 100 × (6400000)²/(8.99 × 10⁹)

Q = 455617.4 C = 4.56 × 10⁵ C

Hope this helps!!!

6 0
3 years ago
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Answer:

0.247 μC

Explanation:

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F_y:  T_y - W = 0\\T_y = m*g = 0.002 kg *9.81m/s^2 = 0.01962 N

T_y = T_*cos(50)\\T = \frac{T_y}{cos(50)} = 0.0305 N

T_x = T*sin(50) = 0.0234 N

The electric force is given by the expression:

F = k*\frac{q_1*q_2}{r^2}

In equilibrium, the distance between the spheres will be equal to 2 times the length of the string times sin(50):

r = 2*L*sin(50) = 2 * 0.1m * sin(50) 0.1532 m

And k is the coulomb constan equal to 9 *10^9 N*m^2/C^2. q1 y q2 is the charge of each particle, in this case, they are equal.

F_x = T_x - F_e = 0\\T_x = F_e = k*\frac{q^2}{r^2}

q = \sqrt{T_x *\frac{r^2}{k}} = \sqrt{0.0234 N * \frac{(0.1532m)^2}{9*10^9 N*m^2/C^2} } = 2.4704 * 10^-7 C

O 0.247 μC

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Answer: a. Frequency. 20, 4, 3, 1, 1, 0, 1

Explanation:

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14000 - 14999 - - - - - - - - - - - - - - - 20

15000 - 15999 - - - - - - - - - - - - - - - 4

16000 - 16999 - - - - - - - - - - - - - - - 3

17000 - 17999 - - - - - - - - - - - - - - - 1

18000 - 18999 - - - - - - - - - - - - - - - 1

19000 - 19999 - - - - - - - - - - - - - - - 0

20000 - 20999 - - - - - - - - - - - - - - 1

Frequency : (20, 4, 3 1, 1, 0, 1)

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